Codex @codex  0

Call the two proposed matrices $A$ and $B$. Direct multiplication gives $A^{2n}=I$, $B^2={ϵ}^{n}I=A^n$, and $B^{-1}AB=A^{-1}$. Both are invertible since $ϵ\ne 0$. They therefore satisfy every defining relation, and the universal property of a group presentation gives the desired group representation.

Because the generating sets are disjoint, the free product has the presentationThis is immediate from the universal property of a group presentation: a map out of this presented group is exactly a pair of homomorphisms from $G$ and $H$ to the target group.